# Is the limit of a sequence of square integrable holomorphic functions also holomorphic?

Suppose we have a convergent sequence of square integrable holomorphic functions $$\{f_n\}$$ defined on a domain $$D$$: they converge to a function $$f$$. Does $$f$$ have to be holomorphic too?

This wikipedia article says that $$f$$ does indeed have to be holomorphic, and that this follows from the fact that for all $$f_n$$, we have $$\sup\limits_{K}f_n(z)\leq C_K\|f_n\|_{L^2(D)}$$ where $$K\subset D$$ is any compact subset of the domain.

How does this prove that the limit of the sequence of functions has to be holomorphic?

• I'm guessing what they showed is $f_n \to f$ in $L^2(D)$ implies $f_n \to f$ locally uniformly, and this implies (by general complex analysis theory) that $f$ is holomorphic – mathworker21 Apr 13 at 1:47
• The question is very imprecise. In what sense does the sequence converge? If the convergence is pointwise and there is no assumption on the boundedness of $L^{2}$ norms then the claim is false. – Kabo Murphy Apr 13 at 5:31

We can prove your estimate by the mean value property for holomorphic functions:

Suppose $$B\subset D$$ is a closed disk with center $$z_0$$ and radius $$R$$. For all functions $$f$$ holomorphic in $$D$$, we have $$f(z_0)= \frac{1}{\pi R^2} \int_{|y|\leq R}f(z_0+y)dA(y)$$, where $$dA$$ is 2 dimensional (area) Lebesgue measure. By Cauchy-Schwarz, we get that $$|f(z_0)|\leq \frac{1}{\pi R^2} \int_{|y|\leq R}|f(z_0+y)|dA(y)\leq \frac{1}{R\sqrt{\pi}} ||f||_{L^2(B)}\leq \frac{1}{R\sqrt{\pi}} ||f||_{L^2(D)}$$

Now let $$K\subset D$$ compact. By compactness, we can find $$B_1, \cdots B_m$$ open disks which cover $$K$$ and whose closures lie in $$D$$. let r be the minimum radius of the disks. Then by our previous computation, we have shown that $$\sup_{z_0 \in K} |f(z_0)| \leq \frac{1}{r\sqrt{\pi}} ||f||_{L^2(D)}$$ which proves that for all $$K\subset D$$ compact, there exists a constant $$C_K$$ depending only on $$K$$ and $$D$$ for which $$\sup_{z_0 \in K} |f(z_0)| \leq C_K ||f||_{L^2(D)}$$ for all functions $$f$$ holomorphic in $$D$$.

It follows that $$L^2$$ convergence of $$\{f_n\}$$ implies that $$\{f_n\}$$ is uniformly Cauchy on each compact subset $$K\subset D:$$ indeed $$f_n-f_m$$ is holomorphic hence $$\sup_{K} |f_n-f_m| \leq C_K ||f_n-f_m||_{L^2(D)} \to 0$$ as $$n,m \to \infty$$ . In particular $$\{f_n\}$$ converges locally uniformly, hence has a holomorphic limit by a standard result in complex analysis (one can prove this, for example by Morera's theorem).

For $$|z-a| < R such that $$\{ |z-a|< T\} \subset D$$ the Cauchy integral formula yields $$f_n(z) = \frac{1}{2i\pi}\int_{|s-a|= R} \frac{f_n(s)}{s-z}ds= \int_0^1 \frac{f_n(a+R e^{2i\pi t})}{1-zR^{-1} e^{-2i\pi t}}dt \\= \frac{1}{T-R}\int_R^T \int_0^1 \frac{f_n(a+r e^{2i\pi t})}{1-zr^{-1} e^{-2i\pi t}}dt dr$$

If $$f_n$$ converges in $$L^2$$ then it converges in $$L^1_{loc}$$ and hence those integrals and their derivatives and power series (obtained from $$\frac{1}{1-zr^{-1} e^{-2i\pi t}} = \sum_{m=0}^\infty z^m r^{-m}e^{-2i\pi mt}$$) converge without problems obtaining that $$f$$ is continuous, $$C^\infty$$, holomorphic, analytic.

• OP does not talk about convergence in $L^{2}$ but the Wikipedia article does talk about it. – Kabo Murphy Apr 13 at 5:33